Posts Tagged ‘work output’

For some time now we’ve been analyzing the helpfulness of the engineering phenomena known as pulleys and we’ve learned that, yes, they can be very helpful, although they do have their limitations. One of those ever-present limitations is due to the inevitable presence of friction between moving parts. Like an unsummoned gremlin, friction will be standing by in any mechanical situation to put the wrench in the works. Today we’ll calculate just how much friction is present within the examplecompound pulley we’ve been working with.

So How Much Friction is Present in our Compound Pulley?

Last time we began our numerical demonstration of the inequality between a compound pulley’swork input, WI, and work output, WO, an inequality that’s due to friction in its wheels. We began things by examining a friction-free scenario and discovered that to lift an urn with a weight, W, of 40 pounds a distance, d1, of 2 feet above the ground, Mr. Toga exerts a personal effort/force, F, of 10 pounds to extract a length of rope, d2, of 8 feet.

In reality our compound pulley must contend with the effects of friction, so we know it will take more than 10 pounds of force to lift the urn, a resistance which we’ll notate FF. To determine this value we’ll attach a spring scale to Mr. Toga’s end of the rope and measure his actual lifting force, FActual, represented by the formula,

FActual =F + FF (1)

We find that FActual equals 12 pounds. Thus our equation becomes,

12 Lbs = 10 Lbs + FF (2)

which simplifies to,

2 Lbs =FF (3)

Now that we’ve determined values for all operating variables, we can solve for work input and then contrast our finding with work output,

It’s evident that the amount of work Mr. Toga puts into lifting his urn requires 16 more Foot-Pounds of work input effort than the amount of work output produced. This extra effort that’s required to overcome the pulley’s friction is the same as the work required to carry a weight of one pound a distance of 16 feet. We can thus conclude that work input does not equal work output in a compound pulley.

Next time we’ll take a look at a different use for pulleys beyond that of just lifting objects.

Last time we began work on a numericaldemonstration and engineering analysis of the inequality of work input and output as experienced by our example persona, an ancient Greek lifting an urn. Today we’ll get two steps closer to demonstrating this reality as we work a compound pulley’s numerical puzzle, shuffling equations like a Rubik’s Cube to arrive at values for two variables crucial to our analysis, d2, the length of rope he extracts from the pulley while lifting, and F, the force/effort required to lift the urn in an idealized situation where no friction exists.

A Compound Pulley’s Numerical Puzzle is Like a Rubik’s Cube

We’ll continue manipulating the work input equation, WI, as shown in Equation (1), along with derivative equations, breaking it down into parts, and handle the two terms within parentheses separately. Term one, (F × d2), corresponds to the force/effort/work required to lift the urn in an idealized no-friction world. It’ll be our focus today as it provides a springboard to solving for variables F and d2.

WI = (F ×d2) + (FF×d2) (1)

Previously we learned that when friction is present, work output, WO, is equal to work input minus the work required to oppose friction while lifting. Mathematically that’s represented by,

WO = WI – (FF × d2) (2)

We also previously calculated WO to equal 80 Ft-Lbs. To get F and d2 into a relationship with terms we already know the value for, namely WO, we substitute Equation (1) into Equation (2) and arrive at,

And because in our example four ropes are used to support the weight of the urn, we know that MA equals 4. We also know from last time that d1 equals 2 feet. Plugging these numbers into Equation (5) we arrive at a value for d2,

d2 ÷ 2 ft= 4 (6)

d2= 4 × 2 ft (7)

d2 = 8 ft (8)

Substituting Equation (8) into Equation (4), we solve for F,

80 Ft-Lbs =F × 8 ft (9)

F = 10 Pounds (10)

Now that we know F and d2 we can solve for FF, the amount of extra effort required by man or machine to overcome friction in a compound pulley assembly. It’s the final piece in the numerical puzzle which will then allow us to compare work input to output.

Last time we performed an engineering analysis of a compound pulley which resulted in an equation comparing the amount of true work effort, or work input,WI, required by machine or human to lift an object, in our case a toga’d man lifting an urn. Our analysis revealed that, in real world situations, work input does not equal work output, WO, due to the presence of friction. Today we’ll begin to numerically demonstrate their inequality by first solving for work output, and later work input.

In our example Mr. Toga lifts an urn of weight, W, equal to 40 pounds to a height, or distance off the floor, d1, of 2 feet. Inserting these values into equation (1) we arrive at,

WO = 40 pounds ×2 feet= 80 Ft-Lbs (2)

where, Ft-Lbs is a unit of work which denotes pounds of force moving through feet of distance.

Now that we’ve calculated thework output, we’ll turn our attention to the previously-derived equation for work input, shown in equation (3). Interrelating equations for WO and WI will enable us to solve for unknown variables, including the force, F, required to lift the urn and the length of rope, d2, extracted during lifting.Once F and d2 are known, we can solve for the additional force required to overcome friction, FF, then finally we’ll solve for WI.

Once again, the equation we’ll be working with is,

WI = (F ×d2) + (FF×d2) (3)

To calculate F, we’ll work the two terms present within parentheses separately, then use knowledge gained to further work our way towards a numerical comparison of work input and work output. We’ll do that next time.

We left off last time with an engineering analysis of energy factors within a compound pulley scenario, in our case a Grecian man lifting an urn. We devised an equation to quantify the amount of work effort he exerts in the process. That equation contains two terms, one of which is beneficial to our lifting scenario, the other of which is not. Today we’ll explore these two terms and in so doing show how there are situations when work input does not equal work output.

Work Input Does Not Equal Work Output

Here again is the equation we’ll be working with today,

WI = (F ×d) + (FF×d) (1)

where, F is the entirely positive force, or work, exerted by human or machine to lift an object using a compound pulley. It represents an ideal but not real world scenario in which no friction is present within the pulley assembly.

The other force at play in our lifting scenario, FF, is less obvious to the casual observer. It’s the force, or work, which must be employed over and above the initial positive force to overcome the friction that’s always present between moving parts, in this case a rope moving through pulley wheels. The rope length extracted from the pulley to lift the object is d.

Now we’ll use this equation to understand why work input, WI,does not equalwork output, WO, in a compound pulley arrangement where friction is present.

The first term in equation (1), (F ×d), represents the work input as supplied by human or machine to lift the object. It is an idealistic scenario in which 100% of energy employed is directly conveyed to lifting. Stated another way, (F ×d) is entirely converted into beneficial work effort, WO.

The second term, (FF×d), is the additional work input that’s needed to overcome frictional resistance present in the interaction between rope and pulley wheels. It represents lost work effort and makes no contribution to lifting the urn off the ground against the pull of gravity. It represents the heat energy that’s created by the movement of rope through the pulley wheels, heat which is entirely lost to the environment and contributes nothing to work output. Mathematically, this relationship between WO, WI, and friction is represented by,

WO = WI – (FF×d) (2)

In other words, work input is not equal to work output in a real world situation in which pulley wheels present a source of friction.

Next time we’ll run some numbers to demonstrate the inequality between WI and WO.

Last time we saw how the presence of friction reduces mechanical advantage in an engineering scenario utilizing a compound pulley. We also learned that the actual amount of effort, or force, required to lift an object is a combination of the portion of the force which is hampered by friction and an idealized scenario which is friction-free. Today we’ll begin our exploration into how frictionresults in reduced work input, manifested as heat energy lost to the environment. The net result is that work input does not equal work output and some of Mr. Toga’s labor is unproductive.

Friction Results in Heat and Lost Work Within a Compound Pulley

In a past blog, we showed how the actual force required to lift our urn is a combination of F, an ideal friction-free work effort by Mr. Toga, and FF , the extra force he must exert to overcome friction present in the wheels,

FActual =F + FF (1)

Mr. Toga is clearly working to lift his turn, and generally speaking his work effort, WI, is defined as the force he employs multiplied by the length, d, of rope that he pulls out of the compound pulley during lifting. Mathematically that is,

WI = FActual × d (2)

To see what happens when friction enters the picture, we’ll first substitute equation (1) into equation (2) to get WI in terms of F and FF,

WI = (F + FF ) × d (3)

Multiplying through by d, equation (3) becomes,

WI = (F ×d )+ (FF × d) (4)

In equation (4) WI is divided into two terms. Next time we’ll see how one of these terms is beneficial to our lifting scenario, while the other is not.

We’ve been discussing the mechanical advantage that compound pulleys provide to humans during lifting operations and last time we hit upon the fact that there comes a point of diminished return, a reality that engineers must negotiate in their mechanical designs. Today we’ll discuss one of the undesirable tradeoffs that results in a diminished return within a compound pulleyarrangement when we compute the length of rope the Grecian man we’ve been following must grapple in order to lift his urn. What we’ll discover is a situation of mechanical overkill – like using a steamroller to squash a bug.

As presented in a past blog, the equations for work input, WI, and work output, WO, we’ll be using are,

WI = F × d2

WO = W × d1

Now, ideally, in a compound pulley no friction exists in the wheels to impede the rope’s movement, and that will be our scenario today. Our next blog will deal with the more complex situation where friction is present. So for our example today, with no friction present, work input equals output…

WI = WO

… and this fact allows us to develop an equation in terms of the rope length/distance factors in our compound pulley assembly, represented by d1 and d2, …

F × d2 = W × d1

d2 ÷ d1 = W ÷ F

Now, from our last blog we know that W divided by F represents the mechanical advantage, MA, to Mr. Toga of using the compound pulley, which was found to be 16, equivalent to the sections of rope directly supporting the urn. We’ll set the distance factors up in relation to MA, and the equation becomes…

d2÷ d1 = MA

d2 = MA × d1

d2 = 16 ×2 feet = 32 feet

What we discover is that in order to raise the urn 2 feet, our Grecian friend must manipulate 32 feet of rope – which would only make sense if he were lifting something far heavier than a 40 pound urn.

In reality, WI does not equal WO, due to the inevitable presence of friction. Next time we’ll see how friction affects the mechanical advantage in our compound pulley.